Let x = cups of 11% wheat, and y = cups of 63% wheat
He wants 12 cups in total, so we need:
x + y = 12
These 12 cups need to have 50% wheat, which means we are looking at 12*.5 = 6 cups of wheat in total, so our second equation is:
.11x + .63y = 6
We have two equations and two variables. So we can solve by using any means you want. I used substitution:
x = 12 - y
.11(12-y) + .63y = 6
1.32 -.11y + .63y = 6
.52y = 4.68
y = 9
Which means x = 3
Thus, 3 cups of 11% and 9 cups of 63%.
It's actually PEMDAS. P is for parenthesis. E is for exponents. M is for multiplication. D is for division. A is for Addition. S is for subtraction.
Here's an example that includes all of these:
(6-2)²x4/2+6-4
P first.
(6-2) = 4
(4)²x4/2+6-4
Now E.
(4)² = 16
16x4/2+6-4
Then M.
16x4 = 64
64/2+6-4
Next is D.
64/2 = 32
32+6-4
Now A.
32+6 = 38
38-4
And finally, S.
38-4 = 34.
I hope this helped
The probability of drawing a green marble will be found as follows:
1. First, find your total number of outcomes by adding together all of the types of marbles to find the total you might draw from the bag so that you may make this a proportion. Now, you get your type of marble you want to know the probability of, and you put the total number of green marbles over this number. The total number of possibilities is 50, and the number of green marbles is 16.
2. I'm not sure what type of answer you are looking for, but the following answers could be the following:
- 16/50
- 32%
-.32
Hope this helps.
Answer:
b = 10
Step-by-step explanation:
a) Solve by inspection. b=10 makes both fractions equal to 1.
Both the numerator and denominator of the fraction on the right are 2 greater than those of the fraction on the left. The fractions will be equal when numerator and denominator are equal.
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b) Multiply by 60 and solve in the usual way.
6b = 5(b+2)
b = 10 . . . . . . . . . . subtract 5b
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c) Expand the fraction on the right, subtract the variable term.
b/10 = b/12 + 2/12
b(1/10 - 1/12) = 1/6
b/60 = 1/6 . . . . simplify
b = 60/6 = 10 . . . . multiply by 60
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<em>Comment on the question</em>
There is no "y", so you cannot solve for y. This is another example of poor quality control (editing) of math curriculum materials. It is my sincere wish that you not have to put up with that. Math is tough enough without this added confusion.